Find points at which the tangent to the curve `y=x^3-3x^2-9x+7`is parallel to the x-axis.

Explanation

Equation of curve
` y = x^(3) - 3x^(2) - 9x +7`…(1)
`rArr (dy)/(dx)= 3x^(2) - 6x-9`
The tangent of the curve will be parallel to x-axis at those point where slope = 0.
`:. 3x^(2) - 6x - 9 =0`
`rArr x^(2) - 2x - 3 = 0`
` rArr(x-3)(x+1) = 0`
` rArr x= 3 or x =- 1`
put x = 3 in equation (1)
` y = 27 - 27 - 27 +7 =-20`
Put x =- 1 in equation (1)
` y =- 1- 3+9+7=12`
`:.` Required point = (3, -20) and ( -1, 12)